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Mashutka [201]
3 years ago
10

This system consists of a marble rolling down a ramp. Assume the mass of the marble is 5.0 g. Decide what would be needed to mea

sure or determine the other quantities to calculate gravitational potential energy and kinetic energy. Solve the equation for the speed of the marble at the bottom of the ramp and compare it to the experimentally measured value.
Physics
1 answer:
Nina [5.8K]3 years ago
5 0

Answer: v = √2.g.h

Explanation:

If we assume that the marble can be approximated by a point mass, and that it starts from rest at a height h, at that moment, all the energy of the system will be gravitational potential energy, that can be written as follows:

U₁ = m. g. h

As we know m, and g is a constant equal to 9.8 m/s², we will need to measure  the height h, either directly, or in an indirect way from the value of the angle that the ramp does with the horizontal, and the measured value of  the distance travelled along the ramp, x.

So, we could write U₁ as follows:

U₁ = m . g. x. sin θ

Now, at the bottom of the ramp, neglecting fricition, all this potential energy must become kinetic energy, as follows:

U₁ = K₂ ⇒ mgh = 1/2 m(v₂)²

Simplifying and solving for v₂ (the speed of the marble at the top of the bottom), we have:

v₂ = √2.g.h

Once the marble has reached to the bottom of the ramp, it has no more net  external forces acting on it (neglecting friction), it must continue moving at constant speed, equal to v₂.

This value can be measured easily, measuring the displacement between 2 points, and the time used to pass between those points, and computing v₂ as follows:

v₂ = Δx / Δt

If the measured value is different to the one calcultated (beyond the expected experimental error) this means that the friction was not so negligible.

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Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?
Hitman42 [59]

The height, h to which the package of mass m bounces to depends on its initial velocity, v and the acceleration due to gravity, g and is given below:

h = \frac{v^{2}}{2g}

<h3>What are perfectly elastic collision?</h3>

Perfectly elastic collisions are collisions in which the momentum as well as the energy of the colliding bodies is conserved.

In perfectly elastic collisions, the sum of momentum before collision is equal to the momentum after collision.

Also, the sum of kinetic energy before collision is equal to the sum of kinetic energy after collision.

Since some of the Kinetic energy is converted to potential energy of the body;

\frac{mv^{2}}{2} = mgh

h = \frac{v^{2}}{2g}

Therefore, the height to which the package m bounces to depends on its initial velocity and the acceleration due to gravity.

Learn more about elastic collisions at: brainly.com/question/7694106

7 0
2 years ago
If you can simply pour sand into a cup then why is it not a liquid?
sleet_krkn [62]
If you, for example, poured it onto a wide cup with a volume equal to the total volume of the sand particles, the sand would not spread out to fill the container but would bunch up together in the middle.
6 0
3 years ago
Read 2 more answers
An infinite conducting cylindrical shell of outer radius r1 = 0.10 m and inner radius r2 = 0.08 m initially carries a surface ch
gayaneshka [121]

Answer:

Explanation:

Solution is in the picture attached

8 0
3 years ago
An electron with speed 2.45 x 10^7 m/s is traveling parallel to a uniform electric field of magnitude 1.18 x 10^4N/C . How much
cupoosta [38]

Answer:

time will elapse before it return to  its staring point is 23.6 ns

Explanation:

given data

speed u = 2.45 × 10^{7} m/s

uniform electric field E = 1.18 × 10^{4} N/C

to find out

How much time will elapse before it returns to its starting point

solution

we find acceleration first by electrostatic force that is

F = Eq

here

F = ma by newton law

so

ma = Eq

here m is mass , a is acceleration and E is uniform electric field and q is charge of electron

so

put here all value

9.11 × 10^{-31} kg ×a = 1.18 × 10^{4} × 1.602 × 10^{-19}

a = 20.75 × 10^{14} m/s²

so acceleration is 20.75 × 10^{14} m/s²

and

time required by electron before come rest is

use equation of motion

v = u + at

here v is zero and u is speed given and t is time so put all value

2.45 × 10^{7} = 0 + 20.75 × 10^{14} (t)

t = 11.80 × 10^{-9} s

so time will elapse before it return to  its staring point is

time = 2t

time = 2 ×11.80 × 10^{-9}

time is 23.6 × 10^{-9} s

time will elapse before it return to  its staring point is 23.6 ns

7 0
3 years ago
You and your family are traveling to daytona beach for a vacation and your dad wants to get there in 7 hours. If daytona is 725
ollegr [7]

Answer:

103.57 Km/h

Explanation:

From the question given above, the following data were obtained:

Distance = 725 Km

Time = 7 hours

Speed =?

Speed can be defined as the distance travelled per unit time. Mathematically, it is expressed as:

Speed = Distance /time

With the above formula, we can calculate how fast he will drive (i.e the speed) in order to get there on time. This is illustrated below:

Distance = 725 Km

Time = 7 hours

Speed =?

Speed = Distance /time

Speed = 725 / 7

Speed = 103.57 Km/h

Thus, to get there on time, he will drive with a speed of 103.57 Km/h

4 0
3 years ago
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